# Definition:Cantor Set

## Contents

## Definition

### As a Limit of Intersections

Define, for $n \in \N$, subsequently:

- $k \left({n}\right) := \dfrac {3^n - 1} 2$

- $\displaystyle A_n := \bigcup_{i \mathop = 1}^{k \left({n}\right)} \left({\frac {2 i - 1} {3^n} \,.\,.\, \frac {2 i} {3^n} }\right)$

Since $3^n$ is always odd, $k \left({n}\right)$ is always an integer, and hence the union will always be perfectly defined.

Consider the closed interval $\left[{0 \,.\,.\, 1}\right] \subset \R$.

Define:

- $\mathcal C_n := \left[{0 \,.\,.\, 1}\right] \setminus A_n$

The **Cantor set** $\mathcal C$ is defined as:

- $\displaystyle \mathcal C = \bigcap_{n \mathop = 1}^\infty \ \mathcal C_n$

### From Ternary Representation

Consider the closed interval $\left[{0 \,.\,.\, 1}\right] \subset \R$.

The **Cantor set** $\mathcal C$ consists of all the points in $\left[{0 \,.\,.\, 1}\right]$ which can be expressed in base $3$ without using the digit $1$.

From Representation of Ternary Expansions, if any number has two different ternary representations, for example:

- $\dfrac 1 3 = 0.10000 \ldots = 0.02222$

then at most one of these can be written without any $1$'s in it.

Therefore this representation of points of $\mathcal C$ is unique.

### As a Limit of a Decreasing Sequence

Let $I_c \left({\R}\right)$ denote the set of all closed real intervals.

Define the mapping $t_1: I_c \left({\R}\right) \to I_c \left({\R}\right)$ by:

- $t_1 \left({\left[{a \,.\,.\, b}\right]}\right) := \left[{a \,.\,. \frac 1 3 \left({a + b}\right)}\right]$

and similarly $t_3: I_c \left({\R}\right) \to I_c \left({\R}\right)$ by:

- $t_3 \left({\left[{a \,.\,.\, b}\right]}\right) := \left[{\frac 2 3 \left({a + b}\right) \,.\,. b}\right]$

Note in particular how:

- $t_1 \left({\left[{a \,.\,.\, b}\right]}\right) \subseteq \left[{a \,.\,.\, b}\right]$
- $t_3 \left({\left[{a \,.\,.\, b}\right]}\right) \subseteq \left[{a \,.\,.\, b}\right]$

Subsequently, define inductively:

- $S_0 := \left\{{\left[{0 \,.\,.\, 1}\right]}\right\}$
- $S_{n+1} := t_1 \left({C_n}\right) \cup t_3 \left({C_n}\right)$

and put, for all $n \in \N$:

- $C_n := \displaystyle \bigcup S_n$

Note that $C_{n+1} \subseteq C_n$ for all $n \in \N$, so that this forms a decreasing sequence of sets.

Then the **Cantor set** $\mathcal C$ is defined as its limit, i.e.:

- $\mathcal C := \displaystyle \bigcap_{n \mathop \in \N} C_n$

These definitions are all equivalent, as shown on Equivalence of Definitions of Cantor Set.

## Also known as

Some sources refer to the **Cantor set** as **Cantor's discontinuum**.

## Also see

- Equivalence of Definitions of Cantor Set
- Cantor Space, the
**Cantor set**endowed with the Euclidean topology.

- Results about
**the Cantor set**can be found here.

## Source of Name

This entry was named for Georg Cantor.